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Section A : Straight Objective Type1. Answer (3)
Due to ortho effect,
COOHOH
is most acidic.
2. Answer (3)On the abstraction of H(d), allyl free radical as
well as tertiary free radical is formed.
3. Answer (3)Electron withdrawing group increases rate for SN
reactions.
4. Answer (2)
Aromatic + resonance stabilised
CH2 resonance stabilised
=
22 HCCHCH resonance stabilised
=
HCCH2 least stable due to presence of positive charge on sp
hybridised carbon.
5. Answer (4)Number of stereoisomers = 2n (n = number of
asymmetric carbon atoms + number of double bonds)
6. Answer (1)Meso tartaric acid is optically inactive due to
presence of plane of symmetry.
7. Answer (1)
The Fisher projection of the compound is HO H
CH3 CH3
OHH
and it has R & R configuration on both
asymmetric C-atom.8. Answer (1)
H C C C C HH
H
H
HH H
p orbitals of carbon being restricted rotation
Organic Chemistry UNIT 2
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Success Magnet (Solutions) Organic Chemistry
9. Answer (3)COOHHOHCHOHCCH3
No. of chiral carbon atoms = 2No. of optical isomers = 2n = 22 =
4
10. Answer (2)
123
4
56 7 Cl
3
3
3
7
Positive charge due to delocalisation of electrons is present 1,
3, 5 and 7 carbon.11. Answer (1)
Bridge head carbocation is not possible.12. Answer (2)
OH
ring expansion
H+
13. Answer (1)N2 is good leaving group, resulting in the
formation of carbene.
14. Answer (4)OCH3 group is +R group which would decrease the
magnitude of + charge.
15. Answer (1)H H
OH
Conjugate base is resonance stabilised and is aromatic.16.
Answer (2)
Compounds containing N and S respond to this test.17. Answer
(2)
HNO3 decomposes Na2S and NaCN present in Lassaigne extract
becuase otherwise they will giveprecipitation with AgNO3.
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Organic Chemistry Success Magnet (Solutions)
18. Answer (3)
Stabilised due to aromatic nature and is resonance
stabilised.
19. Answer (3)
CH = CH O H CH CHO2
O
O
OH
OH
NH NH2
:
20. Answer (4)Pyrolle is least basic as the lone pair is
contributing to aromaticity of the molecule.
In pyridine N
N is sp2 hybridised.
NH
is most basic, therefore 6 member ring is least strained.
21. Answer (2)
CH O CH3 2
: stabilised due to lone pair of O.
22. Answer (2)
B C cyclic delocalisation of 6 electrons.
23. Answer (2)Br is a good leaving group and carbonyl site has
electron deficient carbon.
24. Answer (1)Alkenes undergo electrophilic addition
reaction.Carbonyl undergo electrophilic addition reaction.
25. Answer (4)sp2 hybridised carbon cannot be readily attacked
by nucleophile from back side and attack at bridgeheadcarbon is not
feasible.
26. Answer (2)2 has (4n + 2) electrons and planar ring.
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Success Magnet (Solutions) Organic Chemistry
27. Answer (4)
CH OH2
CH OH2
OHOHHHO
H
H S
S CH OH2
CH OH2
OHOH
HHOH
H S
S
28. Answer (1)
lCBr+
, so Cl will add at most stable carbocationic site.
29. Answer (2)Ketal formation.
30. Answer (4)1 , 2 and 3 has plane of symmetry.
31. Answer (3)In ortho substituted amines, nitrogen moves out of
the plane.
32. Answer (1)1 and 2 are resonance stabilised but 2 is less
stabilised since it has electron withdrawing nitro group.
33. Answer (2)Aliphatic amine is more basic and lone pair e
of pyrolle participate in resonance.34. Answer (4)
Percentage of 1 chloro-3-methyl butane %88.131006.7593
8.3251163110031
=++
=
+++
=
35. Answer (1)
Cl C CH CH3
Cl
Cl
< Cl C CH2CH2
Cl
Cl
(more stable carbocation)
36. Answer (4)H3O+ gives A i.e. Markownikovs addition, through
carbocation formation. B formed by hydroborationoxidation that
gives anti-Markownikovs product.
37. Answer (2)
H+
H
Hydride transfer
H
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Organic Chemistry Success Magnet (Solutions)
38. Answer (2)OH group of molecule will behave as a basic site
in presence of HCl.
39. Answer (2)Gauche form is more stable due to presence of
intramolecular hydrogen bonding.
40. Answer (2)Lemeiux reagent will add OH at each site after
cleavage of every bond.
41. Answer (3)Br
Br 2NaNH2 NaNH2 Na C H Br2 5C H2 5
42. Answer (2)
C C C CCH3
CH3
H
H
C CH3H
H2 CH = CH CCH
CH CH3CH3
(optically inactive)H
CH3
Triple bond is more sensitive for addition H2 than double
bond.43. Answer (3)
CHOCHHCCaCCCaO 3SOHHg
22OH
242
22
++
44. Answer (2)
C = CH
H C3 H
CH3 CH N2 2: CH2 C C
H
H C3 H
CH3
45. Answer (2)Na give electron resulting the formation of free
radical.
46. Answer (2)
COONaSodalime
COCl
Anhyd. AlCl3
CO
A B (ketone)47. Answer (1)
Cl AlCl3
48. Answer (2)
CH CH C CH3 2 H O2Hg , H2+ +CH CH C = CH3 2 2
OHCH CH C CH3 2 3
O
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Success Magnet (Solutions) Organic Chemistry
49. Answer (2)KMnO4 (H+) cold gives syn hydroxylation.
50. Answer (3)
ClCl,
Cl
Cl,
Cl
Cl, Cl
Cl
51. Answer (4)Peracids give epoxy formation.
52. Answer (1)Peroxy acid form epoxy alkane.
53. Answer (2)
CH C CNa3 C H Br2 5 CH C H CH3 2 3 C C54. Answer (2)
OCHOO3
55. Answer (2)Thiophene undergoes electrophilic substitution at
2nd position.
56. Answer (2)
(CH ) C == CH3 2 2 H+ (CH ) C CH3 2 3 (CH ) C == CH3 2 2 (CH ) C
CH3 2 3
CH C (CH2 3)2H+
(CH ) C 3 3 CH3CH == C
CH 3CH 3
57. Answer (2)CH = CH CH = CH2 2 H21, 4 addition CH CH = CH CH3
3
A
O /H3 2O1, 4 addition 2CH COOH3
B(Zinc is not present hence CH3CHO first formed oxidised to
CH3COOH)
58. Answer (3)
CH CH CH == CH3 2 H+
CH CH CH CH3 3CH3
1-2 H shift CH C CH CH3 2 3CH3
Br CH C CH CH3 2 3CH3
Br
(3 more stable than 2)
59. Answer (3)Triple bond is more reactive than double bond for
hydrogenation.
60. Answer (1)Chlorine free radical is prepared in presence of
ultraviolet light.
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Organic Chemistry Success Magnet (Solutions)
61. Answer (3)Carbocation formation and ring expansion takes
place.
62. Answer (1)Due to high electronegativity of carbon.
63. Answer (3)
CH C C CH3 2 CH3O3 CH C C C H3 52
O O
Li AlH4
CH C C C H3 52
OH OH
H HChiral centre = 2Stereoisomer = 4
64. Answer (4)Free radical substitution takes place in presence
of sunlight.
65. Answer (3)
COCl
NO2
Anhyd.AlCl3
CO
NO2
Given product.
66. Answer (1)
1 Kolbes electrolytic reaction.
67. Answer (4)
33CH
3NaNH
3 CHCCCHNaCCCHHCCCH 32
68. Answer (3)Williamsons synthesis.
69. Answer (4)Benzyl fluoride form soluble AgF and alkyl bromide
give yellow ppt of AgBr. Vinyl chloride does not react.
70. Answer (3)Markownikov product.
71. Answer (1)Retention is dominating over inversion.
72. Answer (3)Saytzeff product.
73. Answer (1)Elimination of HCl resulting the formation of
cyclic ether.
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Success Magnet (Solutions) Organic Chemistry
74. Answer (2)
Cl
Cl
NH NH
2
3 ClBenzyne
NH2
Cl
NH2H H2 N NH2 Cl
NH2
.
75. Answer (4)
C H Br2 5
aq KOH
alc
C H OH(1)
2 5
CH = CH2 2 (2)
C2H5OH and CH3OCH3 are isomers
22H
52 CHCHOHHC =+
.
76. Answer (3) NO2 group is electron withdrawing it show I &
R effect. More over it stabilize anion more than other group.
77. Answer (1)Inversion product.
78. Answer (1) NO2 at o and p position stabilize anion
F
NO2
NO2
CH3+ OH
F
NO2
NO2
CH3
OH
F
NO2
O2N
CH3
OH
79. Answer (3)Iodoform test.
80. Answer (1) NO2 group show I & R effect.
81. Answer (4)
+ I2
I
+ HI .
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Organic Chemistry Success Magnet (Solutions)
82. Answer (3)SN1 reaction carbocation formation.
83. Answer (2)
Br
BrAgOH Br Br
OH
.
84. Answer (4)Depends on formation of free radical
Br(1 eq)
2
h
Isomeric monobrominated product
Br2h
85. Answer (4)Carbene formation then ring expansion takes
place.
86. Answer (1)Due to presence of benzylic group
CH Cl2
Cl
NaOH CH OH2
Cl
87. Answer (3)
CH CH SH3 2CH O
CH OH33
CH CH S3 2CH CH22 C
O
CH CH SCH CH O23 2 2H O/H2
+
CH CH SCH CH OH23 2 2 .
88. Answer (3)SN1 involves cyclic formation and
R OS = OCl Cl
S = OO
R+
R Cl + SO2
Where R = CDH
CH3.
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89. Answer (4)
OH
OH
O H
OH
OH
O
Br
90. Answer (4)
CH CH = C3H
COOHBr Cl+
CH CH C COOH3Cl Br
H
4 stereoisomerstherefore2 chiral carbon
91. Answer (2)Chlorination takes place at more probable free
radical site.
92. Answer (1)
CH C CH3 2O
H OH2
18
+H3C C CH2
OH
CH3
H O218
Stable carbocation
C C CH2H OH3
CH3
O18HH
C C CH2H OH3
CH3
O18
H
93. Answer (1)
C)CH( 33 Stable carbocation.94. Answer (3)
N
OH
O
O less stronger than
N OO
OH
Intramolecular H-bonding.
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Organic Chemistry Success Magnet (Solutions)
95. Answer (1)
C
C
O
O
O + 2
OH
H
C
CO
O
OHOH
+ H O2
(phenolphthalein)
96. Answer (1)COOH
+ NaHCO3
COONa
+ CO + H O2 2
Phenol does not react with NaHCO3.97. Answer (3)
R O R + HI R OH + RI 3AgNO AgI ppt.
98. Answer (1)Depends on electronegativity of elements and
stability of conjugate pair.
99. Answer (3)Due to presence of intramolecular H bonding in
(a).
100. Answer (4)Depends on pKa value
pKa acid of Strength1
101. Answer (3)E1 Saytzeff oriented product.
102. Answer (3)Ring expansion
CH OH2H /+ CH2
HH+
103. Answer (3)Both NaBH4 and LiAlH4 reduces acid chloride to
alcohols.
104. Answer (3)
OH
NO2 is weaker acid than H2CO3.
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Success Magnet (Solutions) Organic Chemistry
105. Answer (3)OH
+ FeCl3O Fe O
O+ HCl
(ferric phenoxide)or (C H O) Fe6 5 3
[Fe(C H O) (H O) ]ClSoluble complex
6 5 3 2 3 3
H O2
106. Answer (3)
CH3 KMnO4CH3
OH
OHH
32
CrOCH COOH
3
3
CH3OH
O
107. Answer (3)THF is good solvent for Grignand reagent.
108. Answer (2)CH OH2CH OH2
PCl5 CH Cl2CH Cl2
+ POCl3
109. Answer (4)
OHH+
FTrans (anti addition)
BrCCl
2
4Mesoform
(1)
cis
Br
anti addition2
Racemicmixture
d, forml
Mesoform + d, = 3 stereoisomerl1 2
110. Answer (1)Ether have low boiling point, more volatile
whereas alcohols due to presence of intermolecular H bonding
havehigh boiling point therefore less volatile.
CH OCHCH3 3
CH3
HIS 2N
CH I + HO CH CH3 3
CH3
NaOI
CH COONa + CHI3 3(yellow ppt)
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Organic Chemistry Success Magnet (Solutions)
111. Answer (2)
CH3
O
H+
OH
CH OH2
112. Answer (1)OC OHC O
OH+
HO CH
HO CH2
2
A
O
O O
O
HO CH
HO CH
2
2
H SO2 4
O
O
113. Answer (2)A Base abstract acidic H.
B Carbocation formation due to H+ ion.
114. Answer (1)B is primary alcohol and C is secondary alcohol.
Primary alcohol produces carboxylic acid and secondaryalcohol
produces ketone.
115. Answer (2)
CHClKOH
3
OH OH
CHO
ReimerTiemannreaction
Cannizzaroreaction
50%KOH
OH
COOK+
OH
CH OH2
.
116. Answer (1)
C H CO H6 5 3 OHBr
OH
Br(A) Trans 2-Bromocyclohexanol
117. Answer (4)
Compound -1 CH = CH O 2 CHCH 3
CH 3
H O
H2
+CH CH3 2 3OH + CH C OH
CH3
H
Compound-2 CH = C O 2 C H2 5H O
H2
+CH C OH + C H OH3 2 5
CH3
H
CH3
Both 1 & 2 gives same product.
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Success Magnet (Solutions) Organic Chemistry
118. Answer (1)C H CH 6 5 2 CH OH2 CH2
LiAlH(reduce both C = C andC = O bond)
4
C H CH =6 5 CH CH = Ocrotonaldehyde
NaBH4(does not reduce
C = C bond
C H CH =6 5 CH CH OH2(Y)
(X)
119. Answer (4)CH3OH < CH3 O CH3 < C6H5OH.
120. Answer (3)Jones reagent oxidies 1 alcohol to aldehyde
CH OH2 Jone'sreagent
CHO
In choice (4), tertiary alcohol is resistant to oxidation at
room temperature.
R X + ORaN+
R O R + NaX.
121. Answer (2)Order of boiling point of isomeric alcohols 3
> 2 > 1.
122. Answer (2)
+ O
O
O
O
O
O
123. Answer (4)Cross Cannizzaro reaction
HCHO is always oxidised to HCOOH other part reduced to
alcohols.
124. Answer (4)
C OH + [(CH ) CO] Al3 3 3 > C = O
125. Answer (1)O CHO KOH
50%+
O COOH O CH OH2
no hydrogen atom
Undergo Cannizzaro
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Organic Chemistry Success Magnet (Solutions)
126. Answer (2)
CH CH = CH 3 CH CH3H CrO in aq
acetone (Jones reagent)2 4
OH
CH 3 CH = CH C CH3
O
127. Answer (2)
CHO(CHOH) CH OH
4
2
5HIO4 5HCOOH + HCHO
128. Answer (2)(A) Br is good leaving group and carbocation
stabilize by allylic resonance.
(C) Nucleophilic attack fastest at CO due to presence of H.129.
Answer (1)
NaBH4 reduces only carbonyl compounds.
130. Answer (2)Only acid bromide and alkyl bromide undergo
hydrolysis with alkali Br attached with benzene ring is
resonancestabilize therefore it give substitution at temperature
and pressure.
131. Answer (4)
HO CH CH2NH3
COOH is most acidic so pH is low.
132. Answer (3)
C = O + H+
HO
RR C = O H
OH
R C OH
OH
133. Answer (3)
CH CH CH C N3 2 2CH MgI3 CH CH CH C = NMgI3 2 2
H O3+
CH CH CH C = O3 2 2
CH3
+ Mg IOH
CH3
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Success Magnet (Solutions) Organic Chemistry
134. Answer (4)
CH C Cl3H S2
[H S H]CH C SH + HCl3
O O
135. Answer (4)
CH C O 3 CH3 + CH Mg Br3
O
CH C O 3 CH3
MgBr
CH3
H O2 CH C + 3 MgBr
CH3
O
OH+ CH OH3
(1) CH MgBr(2) H O
3
3+
CH C 3 CH3
CH3
OH
O
136. Answer (2)
H O CH 2 C OH
OSOCl2 Cl CH 2 C Cl
OCH OH3 Cl CH 2 C OCH3
O
137. Answer (3)
CH3 CH2 C NH CH3
O
It should be noted that acid will be neutralised with amine.138.
Answer (2)
keto acid undergo decarboxylation fast due to formation of
resonance stabilised carbanion.
PhCOCH2COOH
PhCOCH3.
138(a). Answer (3) IIT-JEE 2008
Ph
O O
OH
CO2 Ph
CCH3
OI + NaOH 2
PhC
O
ONa + CHI3
GE-keto acid
iodoform reaction
139. Answer (4)HIO4 can oxidise only two adjascent oxygen
containing carbon atoms.Aldehydic and 2 alcohol oxidise to formic
acid whereas 1 alcohol to aldehyde by periodic acid.
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Organic Chemistry Success Magnet (Solutions)
140. Answer (2)
O
O
+ (H Br)
O H
O
Br
OH
OBrH
OHBr
OH
H+
141. Answer (1)Baeyers villiger oxidation.
142. Answer (3)
O
NH OH
H2
+
NOH
Na/C H OH2 5
NH2
H CH3
143. Answer (2)HVZ followed by elimination.
144. Answer (3)Ethylene glycol is protecting group for
carbonyl.
145. Answer (3)(A) NaBH4 reduces cabonyl group.(B) HBO changes
alkane to alcohol. (Anti-Markownikovs product)
146. Answer (4)Pinacol - pinacolone rearrangement.
147. Answer (1)
O O
O
Stable aromatic
O
Stable aromatic
(1)
(2)
(3)
Anti - aromatic
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Success Magnet (Solutions) Organic Chemistry
148. Answer (4)
NBS Mgdry ether
Br MgBr
CH CN3
H C C = N MgBr3
H O2
H C C = O3
149. Answer (1)Br
Red P2CH CH COOH3 2 CH CHCOOH3
Br
NH3 CH CHCOOH3
NH2
150. Answer (1)
NH2
+ Cl C C Cl
HO
H
H N C CH Cl2
O
AlCl3
N
CH2
H
C = O
H
orON
151. Answer (1)Coupling occurs prefenertially in the para
position to the hydroxyl group. But it this position is blocked
thencoupling occurs at ortho position.
152. Answer (3)
NO2Br /FeBr2 3
NO2NaNO
HCl2
N2Cl
Br 280 K(X)Br
(Z)
Cu powderCl
Br(W)MgEther
Cl
MgBr
C H COCHH O
6 5 3
3+
Cl
C H6 5OH
CH3
(F) (A)
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Organic Chemistry Success Magnet (Solutions)
153. Answer (2)
NH2NaNO
HCl2
N2ClOH
phenanthrene
154. Answer (1)
KCN
OH Cl OH CN
SOCl2
Cl CN
H /Pt2
Cl CH2NH2
NaOHHCl N
155. Answer (2)
NH3
O OH O NH2A
KOHBr2
NH2B
CHClKOH
3
NCC
H O3+
NH2
+ HCOOH
DE
(B & D are same]
156. Answer (1)
NaNO /HCl0 5C
2
NO2
F
NH2
NMe2
N Cl2
NMe2
H /Ni2
NH2
NMe2
157. Answer (2)Hoffmann elimination takes place, resulting the
formation of less substituted alkene.
158. Answer (1)
CH C H3
O
gives iodoform (haloform) test.159. Answer (3)
CH CH COOH3 2SOCl2 CH CH COCl3 2
(CH CH CH ) NH3 2 2 2 CH CH 3 2 C N
OC H3 7
C H3 7
LiAlH4 (C H ) N3 7 3A CB
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160. Answer (2)Less substituted alkene is more stable
N CH3
CH3CH3
OH H O2 N
CH2
CH3CH3
+ H O2
161. Answer (3)
C NH + O = C C CH 2 3CH3
H3C H CH3
H H
C N = C C CH3CH3
H3C H CH3
H H
CH C N CH CH(CH )3 3 22H
CH3 H
H /Pt2
162. Answer (3)III - 2 lone pair of electrons present on 2 N
atom.IV - Presence of SO2NH2 group.
163. Answer (4)Due to presence of +I effect of methyl group.
164. Answer (2)Formation of quaternary ammonium salt.
165. Answer (1)
O + H +OH
NH
OH
NH
OH2NH
N
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Organic Chemistry Success Magnet (Solutions)
166. Answer (3)Only benadryl is tertiary amine
Choline - quaternary ammonium salt
Benzedrine - primary amine
Coniine - secondary amine.
167. Answer (2)2 amine react with Hinsberg reagent which forms
insoluble material which does not soluble in NaOH.
1 amine react with Hinsberg reagent and finally form soluble
complex. 3 amine does not react with Hinsbergreagent.
168. Answer (2)
CH I3
N H
N CH3CH3
I
AgOH
NCH3
CH3I
AgOH
N CH3CH3
OHN
CH3
CH3
CH I excess3
CH3
169. Answer (2)
N H O2 2
+ NH OH2
170. Answer (1)
C
O
O
O
+ NH3C
O
C
O
OH
NH2
171. Answer (3)
O
C NH2H /Pd.
high pressure2 CH NH2 2
H /Pd.2 CH NH2 2
H2 reduces CONH2 and then benzene ring.
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172. Answer (3)
22OH CORNHOCNR 2 + == .
173. Answer (2)Nitroglycerene is formed by reaction of glycol
(alcohol) and nitric acid (acid).
174. Answer (4)Secondary amine has + I effect.
175. Answer (2)NaBH4 reduces only carbonyl compound where as
LiAlH4 reduces carbonyl and cyanide group.
176. Answer (3)
NH2O
CH3
NH3 OH
177. Answer (1)
O
O
HNN
H
OHH
O
OH
O
O
NN
H
OHH
O
Anion (conjugate pair) stabilize by resonance.178. Answer
(1)
323OH
223LiAlH
3 HNCHCHNHCHCHNCCH 34
+
.
179. Answer (2)
C = O + H N NH C NH 2 2 2
OCH3
Hsemi carbazide
Pyridine
C = N . NH C NH2H3C
H
O
Semi carbazone
NH2 of
O
C NH2 does not undergo condensation due to its resonance
stabilization.
O
C NH2
O
C = NH2
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180. Answer (3)1 amine oxidise to nitro group and 2 alcohol
oxidise to ketone with cold KMnO4.
181. Answer (4)NH OH
H2
+ O H
Et OH
OHH2NOH OH
N OH
H H
N OH
H
N O H
O
H O2
182. Answer (2)NaOH react with carboxylic acid.
183. Answer (4)3 amine cannot form amine oxide.
184. Answer (3)Due to stability of carbocation.
185. Answer (2)
CH CH3 2 C C = O
H
CH3
H
+ H NCH2 3
CH CH3 2 C C = N CH3H
CH3 H
H /Pt.2
CH CH3 2 C C N CH3H
CH3 H H
H
186. Answer (3)Nylon has strongest intermolecular forces (i.e.
hydrogen bonding) out of these.
187. Answer (1)Artificial silk is a polysaccharide while the
other three are polyamides.
188. Answer (1)
n CH2 = CCH3
CH3C CH C CH2 2
CH3
CH3
CH3
CH3
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189. Answer (4)
Teflon FCF
FCF n
is fully fluorinated polymer.
190. Answer (3)
Amino acid are amphoteric in nature. The acidic properties are
due to amino group
3HN .
191. Answer (1)Alanine is NH CH COOH2
CH3
NH CH COOH3
CH3(At low pH)i.e. acidic pH NH CH COO2
CH3(At high pH)i.e. basic pH
192. Answer (1)During vulcanization of natural rubber S S
crosslinks are formed which make it more elastic.
193. Answer (3)
n CH = CH22343 K 373 K, 6 7 atm
Ziegler Natta catalyst[TiCl + (C H ) Al]4 2 5 3
( CH CH )2 2 n
194. Answer (3)The -helix is known as 3.613 helix since each
turn of the helix has approximately 3.6 amino acids and 13membered
ring is formed by hydrogen bonding.
195. Answer (4)Both insulin and carboxy peptidase contain
Zinc.
196. Answer (3)Cashmilon is polyacrylonitrile.
197. Answer (4)
Proline is heterocyclic amino acid. H N H
COOH
amino acid have a primary amino group except proline.198. Answer
(3)
Acrilan is PAN (Poly acrybonitrile).199. Answer (4)
Caprolactum is monomer of Nylon-6 or Perlon N
HO
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200. Answer (1)
Bakelite is made up of
OH
(phenol) and HCHO (formaldehyde)
OH
+ HCHO
OHH C2 CH2
CH2
201. Answer (1)Methoxy group is attached with carbonyl
oxygen.
202. Answer (2)There is a hydroxyl group and an ether attached
to the same carbon forming a hemiacetal.
203. Answer (1)OH group present on right side
H OHCH OH2
204. Answer (1)Fat 130 ATP, carbohydrate 38 ATP, protein 5
ATP.
205. Answer (1)
HCH OH2
OHOH
H
OH
H O
OH
H D glucopyranose
206. Answer (2)
CHO
CHOH
(CHOH)3CH OH2
C H NNH6 5 2H O2
CH = NNHC H6 5CHOH
(CHOH)3CH OH2
C H NNH6 5 2C H NH , NH6 5 2 3
CH = NNHC H6 5C = O C H NNH6 5 2
H O2
CH = NNHC H6 5C = NNHC H6 5
CH OH2
(CHOH)3
glucosozoneCH OH2
(CHOH)3
207. Answer (2)Only phospholipids form bilayer memberane.
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208. Answer (2)
C H O + H O12 12 11 2H+
invertase D(+)-glucose + D()-fructoseC H O6 12 6 C H O6 12 6
( ) = 52.7 D ( ) = 92.4 Dsucrose( ) = +66.5 D
invert ( ) = 19.9 D209. Answer (3)
C12H22O11 + 18[O] conc. HNO3 COOH
COOHoxalic acid
+ 5H O2
210. Answer (2)The excess glucose present in the blood is
transported in the tissue where insulin converts it into
glycogen.The deficiency of insulin causes diabetis.
211. Answer (2)Proteins give ninhydrin and molisch test.
212. Answer (3)
C H O + H O12 22 11 2( ) = +66.5 D
H+
invertase C H O6 12 6D(+)-glucose( ) = 52.7 D
+ C H O6 12 6D()-fructose( ) = 92.4 D
invert sugar ( ) = 19.9 D213. Answer (1)
Glucose and fructose are functional isomers.214. Answer (1)
Glucose and fructose have similar configuration on C3, C4 and C5
so they form same osazone. In osazoneonly C1 and C2 are
involved.
215. Answer (4)-D-glucose and D-glucose are anomers.
216. Answer (3)
C N
O H peptide bond of protein form Hbond.
217. Answer (2)Stearic acid, Palmatic acid and Oleic acid are
higher fatty acids. Their Na salts are used as soaps.
218. Answer (2)Chemical name of vitamin E is tocopherol.
219. Answer (3)Purines contain two rings in which each ring
contains 2 nitrogen atoms.
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220. Answer (4)Denaturation of protein converts quaternary,
tertiary and secondary structure to primary structure.
221. Answer (2)Invertase for carbohydrates, urease for urea and
trypsin for protein.
222. Answer (4)
pH = 1 (acidic). It accepts a proton and exist as cation 3HN is
acidic group.223. Answer (3)
Amide linkage ( CONH2)224. Answer (4)
5 Chiral carbon present in Dglucose H
CH OH2OH
OH
H
HO
H O
OH
HH*
*
*
*
*
225. Answer (3)
H C = O
(CHOH)4CH OH2
HCN H C OH
(CHOH)4CH OH2
H O3+
COOH(CHOH) CH OH
5
2
HI
COOH(CH ) CH
2 5
3
Heptanoic acid
CN
226. Answer (3)
CH OH
C= O
CH OH
2
2
No chiral carbon
No stereoisomers.
227. Answer (2)Ruff degradation.
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Section - B : Multiple Choice Questions1. Answer (1, 2)
In cis form same groups are present on the same side of double
bond.
2. Answer (1, 2, 3, 4)
In +
and
+
aromaticity is maintained in canonical forms, while benzyl
carbocation looses its
aromaticity in canonical forms.
(CH3)3+
C is more stable than benzyl carbocation due to 9
hyperconjugative bonds.
+
CH2 is more stable than benzyl carbocation due to bent p-atomic
orbital.
3. Answer (2, 3)If one + I and one I effect group is present on
both doubly bonded carbon atom then the dipole moment oftrans form
is more than cis form.
4. Answer (1, 2, 4)Beleistein test is not given by those
compound in which F atom is present. Urea and thiourea also
giveBeleistein test.
5. Answer (1, 3)
COOHOH
is more acidic than HCOOH due to ortho effect while
COOH
and
COOH
CH3
are less acidic
than HCOOH.
6. Answer (1, 2)
OH
C O
C
C
C
O
O
H
H
C O
C
C
O
H
HOOC H
,
Conjugate baseof maleic acid
Conjugate baseof fumaric acid
(stability)
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7. Answer (1, 3)
F SbF5
F
Aromatic systemsSbF5
8. Answer (1, 2)Triple bond is more prone to hydrogenation.
9. Answer (1, 2, 4)Cyclooctatetraene has no planar
structure.
10. Answer (1, 2, 3, 4)C4H10O can form 4 alcohols & 3 ethers
which can exhibit chain, position, functional & metamerism.
11. Answer (2, 4)(2) and (4) contain chiral carbon so are
optically active.
12. Answer (1, 2, 3)A, B, C, D are isomers of same compounds
where (A) and (B) make, enantoimeric pair and (C), (D) makeanother
enantiomeric pair.
13. Answer (1, 2)
H
COOH
CH3
H N2
OHH
S
R23
OH
COOH
H
OHH
S
R
23
COOH14. Answer (2, 3, 4)
(1) molecule contains 3 (1 H), 8 (2 hydrogens) and 3 (3
hydrogens).15. Answer (1, 3, 4)
OO has no acidic hydrogen to participate in enol formation.
16. Answer (1, 2, 3)Bromination takes place at most stable free
radical.
17. Answer (1, 2, 3)
ROH + CH MgI CH + Mg3 4I
OR0.37
xx = 74
112 cc22400
CH C CH OH3 2
H
CH3
H+
CH C = CH3 2CH3
O3 CH C = O + HCHO3CH3
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18. Answer (1, 3)A Presence of benzylic H
B Absence of benzylic H
19. Answer (1, 3)
C5H12 + Cl2 H C C CH Cl3 2
CH3
CH3
NaWurtz reaction H C C CH CH C CH3 2 2 3
CH3
CH3
CH3
CH3(B) (C)
20. Answer (1, 2)
O
O OO
CHOC
O
H
C
O
H CHO
O3HOH
OH
OO
H
21. Answer (1, 2, 3)Wurtz reaction is a type of free radical
reaction.
C2H5 Cl2NaCl ++ C2H5 etherdry 2NaClHCbutane104+
Ethyl free radical can disproportionate to give ethane and
ethene.
Ethane62
Ethene4252 HCHCHC2 +
22. Answer (1, 2, 4)
On applying selectivityreactivity principle, only CH CH CH CH3
3CH3
Cl has about 35% in the mixture. Others
have less than 35% in the mixture.
23. Answer (1, 3, 4)2-methyl 2-butene is more stable than
but-2-ene due to presence of 9 hyperconjugative bonds,
whilebut-2-ene contains 6 hyperconjugative bonds.Buta-1, 3-diene
and 2-methyl buta-1, 3-diene are more stable than but-2-ene due to
conjugation.
24. Answer (1, 2, 4)In all molecule, one H2 molecule is added
but only (1), (2) and (4) can give symmetrical diketone on
reductiveozonolysis, while (3) give only ketone.
25. Answer (1, 3, 4)In presence of 1% alkaline KMnO4, C6H5CO3H
and OsO4/OH, syn addition occurs while in presence ofHCO3H, anti
addition occurs.
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26. Answer (1, 3)A by E2 elimination, B by E1 elimination.
Alc. KOH will undergo dehydrohalogenation by anti
elimination.27. Answer (1, 4)
NBS characteristic reagent for allylic brominationBr ortho para
directing
28. Answer (1, 2)
NBS
Br
(A)29. Answer (2, 3)
Due to presence of unsaturation.30. Answer (1, 2, 4)
Clanhyd AlCl3 hydride shift
more stablecarbocation
31. Answer (2, 3)
will give only 1 mono halo substituted derivative.
will give 3 mono halo substituted derivative.
32. Answer (1, 3)SO2Cl2 and Cl2 replaces benzylic H.
33. Answer (2, 3)
C3H7Br + AgNO2 productromin
73major
273 ONOHCNOHC +
due to ambident nucleophilic nature of nitro group.
34. Answer (1, 3)
LiAlH4
CH2
H C3 Br
CH3
CH3(A)
LiAlH4
CH3H C2
Br CH3H C3
(B)
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35. Answer (1, 2, 3, 4)All are feasible reaction
(1), (2) substitution as well as elimination(3) Ring
expansion(4) Ring contraction
36. Answer (1, 2, 3, 4)1-4 addition dominates at high
temperature and with excess of HBr both double bond give addition
product.
1-2 addition dominates at low temperature
37. Answer (1, 2, 4)Br
BrBr
HNO3
Br
BrBr
NO2
Highest melting point 1, 3, 5 tribromobenzne.
38. Answer (2, 4)
(2) AgBrROROAgBrRdry
2 ++
(4) CH3CH2OH K413SOH 42
CH3 CH2 O CH2 CH3
39. Answer (1, 2, 3)In (1), (2) and (3) options, stable
carbocation is formed as an intermediate, so, these give SN1
mechanism,while CH3Cl mainly gives SN2 mechanism because it is 1
alkyl halide and produce less stable carbocation.
40. Answer (1, 3)(1) and (3) on E2 elimination give an alkene
but (2) and (4) cant give an alkene
41. Answer (2, 3)3 alkyl halide mainly gives SN1 mechanism.
Polar protic solvent favours SN1 mechanism while polar
aproticsolvent favours SN2 mechanism.
42. Answer (1, 3, 4)In (2) option, alkene is formed as major
product while in other options, ether is formed as major
product.
43. Answer (3, 4)CH3CH2CH2OH + SOCl2 Pyridine CH3CH2CH2Cl + SO2
+ HClCH3CH2CH2OH + PCl5 CH3CH2CH2Cl + POCl3 + HCl
44. Answer (1, 2, 3)(1) Yellow ppt of AgI(2), (3) Iodoform CHI3
formation
45. Answer (1, 3)(1) Friedel Craft alkylation(3) Friedel Craft
decarbonylation
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46. Answer (1, 2)
OH
H+
H O2+
major minor product47. Answer (3, 4)
(2) RNH2 + CHCl3 + alc.KOH RNC + KCl + H2O
(4)
NH2
+ CHCl3 + alc.KOH
NC
+ KCl + H2O
48. Answer (1, 2, 4)1, 2, 4 gives Hoffmann product(1) Hindered
base(2) EICB (stable conjugate base)(4) Hoffmann elimination
49. Answer (2, 3)
C N
O H
is o, p directing due to N
H
group.
50. Answer (1, 3)Only 1 amine gives carbyl amine test.
51. Answer (1, 2)cis and trans isomers C = C PhH
PhH & C = C
PhHPh H by elimination.
52. Answer (1, 2, 3)
CH Br2
H C5 2
1
2
34
Br
CH2
HOHH+
+CH OH2
H C5 2
CH2
H C5 2
CH2
H C5 2
HOHH+
CH2
H C5 2
OHCH2
H C5 2OH
(1)
(2)
(3)
HOHH+
1
H C5 2
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53. Answer (2, 3, 4)Br
C H ONa2 5
54. Answer (2, 3, 4)
OH
HI/P
55. Answer (1, 2)Water favours the formation of a polar
compound.
56. Answer (1, 4)
Secondary alcohol containing CH
OH
CH3 linkage can give all given three reactions.
57. Answer (2, 4)
O OH+16
16
H
+
OH18 H O2
18
+ OH16
+
58. Answer (1, 2, 3, 4)In (1), (2) and (3) intramolecular
hydrogen bonding is present.
59. Answer (1, 2, 3)
Iodoform test is given by alcohols having the group R CH CH3
OH
. Thus, all the three except (4) give this test.
60. Answer (1, 3)
CH CH C CH3 2 (i) NaNH2(ii) C H Br2 5 CH CH C C C H3 2 2 5
H2Pd/BaSO4
C = CC H2 5
H
H5C2
H
alk.KMnO4
C H2 5
C H2 5
OHOH
HH
(X)
(Y)
optically inactive (Z)
cis alkene
meso compound
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61. Answer (1, 3)CH3COCl and (CH3CO)2O can be used.
62. Answer (3, 4)Both (3) and (4) are stable due to
intramolecular hydrogen bond formation
Cl C CH
Cl OH
Cl OH
and F C CH
F OH
F OH
No hydrogen bonding occurs with Br and I due to large size and
lesser electronegativity.
63. Answer (1, 3)
C H CHCH6 5 3
OH
and CH CH CH3 3
OH
gives iodoform test due to presence of CH3 group adjacent to OH
group.
64. Answer (1, 3)In option (1) and (3), most stable carbocation
is formed as an intermediate.
65. Answer (1, 2)
OHNO2
and
OHNO2
NO2
cant give effervescences with NaHCO3 but
OHNO2
NO2
O N2 can
give effervescences with NaHCO3 due to more acidic nature.
66. Answer (2, 3)
HO C CH3
O
and H N C CH2 3
O
cant give iodoform on warming with NaOH and iodine due to
resonance.
67. Answer (1, 2, 4)Pyroligneous acid cotnains 9 10% CH3COOH, 2
4% CH3OH, 1 2% acetone and rest is water.
68. Answer (1, 2)In esterification, H+ of alcohol reacts with OH
of carboxylic acid to form H2O and the reactivity of carboxylicacid
is 1>2>3. Other statements are correct.
69. Answer (1, 2, 4)
OH
NO2 NO2
NO2
,
COOH
,
SO H3
are stronger acid than H2CO3 acid.
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70. Answer (2, 4)
D OH
HHCH = CH CH CH3
OHK Cr O2 2 7
H+
D O
H C OH
O
+ HO C C CH3SOCl2
Cl C C CH3
OO
+
D
H C Cl
O
(Y)(X)
OO
O
71. Answer (1, 2)
O OCH3
H+
H O2 HOOH OH
+ CH OH3H O2
OOH H
72. Answer (1, 3)(3) does not contain hydrogen and (1) contains
vinylic hydrogen which cannot be easily removed.
73. Answer (3, 4)
O
O
OHconc.
COOOH
Cl
COOC H2 5
CCl3
Br
aq. KOHaq. KOH
74. Answer (1, 2, 3, 4)
R RCHO COOAgNO /NH OH3 4
H C OH
OAgNO /NH OH3 4 HCO3
NHOH NO
AgNO /NH OH3 4
(R can be CH3 C6H5)
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75. Answer (1, 3, 4)CH
CH MgBr2
MgBr
CH2 CH2 OMgBr
OHB H2 6
HCHO
CH2
H2O /OH2
+O
H O /OH2 2
76. Answer (1, 2, 3, 4)O
O
O Phenol Phenolpthaline
ResorcinolCatachol
Nitration
FluorisienAlizarin
2,4 D.N.P.
77. Answer (2, 3)OMDM gives MarkownikovsHBO gives
Anti-Markownikovs productNo, carbocation formation No,
rearrangement takes place
78. Answer (1, 2, 3)
(1) H C CH CH CHO3 2
CH3NaOH CH C C C CHO3
CH3 H H
C H
CH3H3COHH
aldol
(2) H C CH CH C = O + CH MgBr3 2 3
CH3 HH O3
+
CH CH CH C CH3 2 3
CH3 OH
H(3) 3-pentanone and 3 methyl butanol have same molecular
formula(4) On Wolff Kishner reduction it gives 2 methyl butane.
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79. Answer (1, 2, 3)
(1) PhCH MgBr + CH C H2 3 CH C CH Ph3 2OH
H
O
(2) CH CH CH + PhLi2 3 H O3+
OPhCH C CH2 3
OH
H
(3) Ph C C H + CH MgBr3
OH
H
H O3+
PhCH C CH2 3
OH
H80. Answer (1, 2, 4)
(1)
CH3
(2)
(4) CH3
CH381. Answer (2, 3, 4)
(2) NaBH4 reduces carbonyl to 2 alcohol.(4) Silver mirror test
given by aldehydes.
82. Answer (1, 2)N Cl2
Cu/KCN
CN
H O3+
COOH
NaOHCaO
(X) (Y) (Z)83. Answer (1, 3)
(1) H C CH3
OH
(O)KMnO4
C CH3
O
(3) + CH COCl3AnhyAlCl3
COCH3
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84. Answer (2, 3, 4)CH3CHO is less reactive than HCHO but more
reactive than ketones.
85. Answer (2, 3, 4)Ethyl alcohol is less acidic than
phenol.
86. Answer (1, 3)
Sulphur and quinoline N
87. Answer (1, 2, 3)In (4) 2 alcohol is formed.
88. Answer (1, 2, 3)NaHSO3 is used to separate ketone and
aliphatic methyl ketone as well as aldehyde and aliphatic
methylketone.
89. Answer (1, 2, 3, 4)
In (3) COOH group is reduced in preference to C = O group.90.
Answer (1, 2, 3)
Active methylene compounds are used in their reaction.91. Answer
(1, 2, 3, 4)
(1) Cyanohydrin formation(2) Aldehydes give silver mirror(3)
Cross aldol(4) Condensation with ammonia derivative
92. Answer (1, 2, 3, 4)(1) C2H5COOH(2) CH3CH2COOH
(3) )B(
23)A(
223 COOHCHCHOHCHCHCH
(4) CH3CH2COOH93. Answer (1, 2, 3)
CH3Cl
CH3
ClCH3
Cl
KNH2
KNH2/
CH3 CH3
NH2
CH3 CH3NH2
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Success Magnet (Solutions) Organic Chemistry
94. Answer (1, 4)NO2 NO
Fe/H O(v)2NH2
OH
strong acidelectrolysis
95. Answer (1, 4)
R RNH2 OHHNO2
But exceptionally methylamine forms ether.
96. Answer (1, 2, 3, 4)In all given compounds lone pair of N is
resonance stabilized.
97. Answer (1, 2, 4)3 amine has no H atom on N.
98. Answer (1, 3)Nylon66 = [ NH (CH ) NH CO (CH ) CO ]2 6 2 4
n
PMMA = CH C 2
CH3
COOCH3 n99. Answer (1, 2, 3, 4)100. Answer (1, 2, 3)
Fact
101. Answer (2, 4)Fact
102. Answer (1, 2, 4)103. Answer (1, 3, 4 )104. Answer (2, 3,
4)
Sucrose is a disaccaharides of glucose and fructose.105. Answer
(1, 3, 4)
Nylon-66 is a copolymer monomers are hexamethylenediamine and
adipic acid.
106. Answer (1, 2, 4)Natural silk is a polyamide.
107. Answer (1, 2, 4)
CH CH + HCN Ba+2 polymerisation (PAN) CH CH 2
CN n
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Organic Chemistry Success Magnet (Solutions)
108. Answer (3, 4)Fructose is equilibrium with glucose in
alkaline solution and hence gives positive test of aldehydic
group.
109. Answer (2, 4)At pH > 7, H+ from carboxylic acid will be
lost.
110. Answer (1, 2, 3)Glycine is only amino acid which is
optically inactive, since then is no asymmetric centre.
111. Answer (1, 2, 3, 4)112. Answer (2, 3, 4)
Guanine is purine.113. Answer (3, 4)
Remini and Schyrver are tests for formaldehyde.114. Answer (3,
4)115. Answer (1, 2, 3, 4)
Section - C : Linked ComprehensionC1. 1. Answer (4)
HH
sp3 hybridised
sp3 hybridised
So non planar (so not aromatic).
(I)
4 electrons (so not aromatic).
2. Answer (4)
NH
E+
NH
NH
HE
E+
NH
NH
HE
+ E+
NH
NH
HE
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3. Answer (2)
O+
O+
anti aromatic
C2. 1. Answer (4)+ 1 group stabilises free radical.
2. Answer (2)Allyl carbocation is most stable.
3. Answer (1)Stability of carbanion increases with the presence
of electron withdrawing group, whereas decreases withthe presence
of electron releasing group.
C3. 1. Answer (1)
H H
O OH
Aromatic, hence very stable.2. Answer (2)
In (1), (3) & (4) the group attached to CO, involves, CO
in resonance.
O O
Cl
O O
Cl3. Answer (4)
O O O OH
stabilised due to intramolecular hydrogen bonding.C4. 1. Answer
(2)
Electron releasing group decreases acidic nature hence highest
pKa.2. Answer (2)
In aqueous solution 2 amine is more basic than 3 & then
comes 1.
3. Answer (4)For iodine + R is least.
C5. 1. Answer (1)
++= cos2222
1
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Organic Chemistry Success Magnet (Solutions)
2. Answer (2)Symmetrical molecules have zero dipole moment.
3. Answer (1)CCl4 has zero dipole moment. As the number of
chloride atoms increase dipole moment decreases.
C6. 1. Answer (1)CH2
no. of hyperconjugative structures = 3 2 1Alkyl group make the
ring electron rich by their tendency to make ring electron rich by
hyperconjugation.
2. Answer (2)(3) is most stabilised by higher + 1 effect of
isopropyl group.
3. Answer (3)
CH = CH2 H+ methylshift
C7. 1. Answer (4)[O]
O , H O3 2 2COOHCOOH
2. Answer (3)
CH = CH2 2MCPBA
OCH MgI3
OMgI
CH3H O3
+
OH(A) (B)
3. Answer (4)
CH CHO + 3
O+ HCHO
C8. 1. Answer (1)Diels alder reaction involves (4 + 2)
cycloaddition.
2. Answer (2)
HH
C = C C = CH H
H
H
H+ H
H
Br
Br
Br
BBr
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3. Answer (3)
+
B E
O /H O3 2 COOHCOOH
C9. 1. Answer (1)a
is most substituted double bond, so most stable.
2. Answer (3)The structure has OH group on most substituted
carbons of the double bonds because the water attacksthe
carbocation that forms. The most stable carbocation is the one on
the most substituted carbon.
3. Answer (1)All of the double bonds have two substituents. If
on the one end of the bond both the substituents aresame.
Consequently there is no possibility for geometric isomers.
C10. 1. Answer (4)Alkenes are more reactive towards
electrophilic addition reactions but when product formed is
conjugateddiene alkynes give this reaction first.
2. Answer (1)Alkenes are more reactive for electrophilic
addition reaction.
3. Answer (1)Alkenes are more reactive for hydrogenation.
C11. 1. Answer (4)Cis alkene + Anti addition Racemic
mixture.
2. Answer (4)Product produced has no chance of having plane of
symmetry.
HCH3
OHCH3
3. Answer (4)OsO4 Syn hydroxylationH2 Syn additionKMnO4 Syn
hydroxylationBr2/CCl4 Anti addition
C12. 1. Answer (2)
ClCl
(d) + ( )l
Cl
(d) + ( )l
Cl
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2. Answer (2)
Cl probability = 1 3 = 3
Clprobability = 2 3.8 = 7.6
Clprobability = 1 5 = 5
Clprobability = 6 1 = 6
3. Answer (3)Bromination takes place at most stable free radical
and free radical at bridge head carbon is least stable.
C13. 1. Answer (3)Alkoxide ion is a better base and so will
favour E2.
2. Answer (3)Same.
3. Answer (2)Vinyl halide and aryl halide have partial double
bond character between carbon and halogen. in aniline, NH2is a
strong base and hence very weak leaving group, so cannot undergo SN
reaction
C14. 1. Answer (4)
OCH3 is a better base.
2. Answer (1)SN1
3. Answer (3)E1
C15. 1. Answer (3)
( ) n 56n 42M(w) vesderivati acetyl of .tW(w) alkali used of
.tW
=+
M molecular mass of alcohol.
2. Answer (1)Secondary alcohol (glycol) gives formic acid.
3. Answer (1)Two 1 alcohol gives formaldehyde and C = O changes
to CO2.
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C16.
OH
NaOH
ONa
(i) CO4 - 7 atm P
(ii) H
2
+
OH
COOH
A B
NaOHCaO
OH
CBr2(CH CO) O3 2
O COCH3
COOH
NaOH OH
COONa + CH COOH3
OHBrBr
Br D
G H
E
1. Answer (2)2. Answer (1)3. Answer (3)
C17. 1. Answer (2)
BrMg
ether MgBrCH CHO3 CHCH3
OH
H+
CHCH3Ring
expansionBr
CH3
Br
2. Answer (2)
CC H3 7
H3C
OH
Ph
There would be a racemic mixture since carbonyl group is planar
and can attack from both side.3. Answer (3)
Alkyl group (carbanion) attack on carbon atom of carbonyl
group.C18. 1. Answer (2)
Compound B is C7H6O3
O H
C = OOH
H-bond
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Organic Chemistry Success Magnet (Solutions)
2. Answer (1)
OH
COOH
OCOCH3COOHCH COCl3
H+
PhOHH+
(C)
OHCOOPh
(D)
(aspirin)
(salol)
3. Answer (2)Compound D is salol.
C19. 1. Answer (3)
CH3
OH
B
O CH3
A =
2. Answer (3)O COCH3
YCOOH
, Aspirin used as antipyretic.
3. Answer (1)
CH3
OH O CH3
& C H O7 8
C20. 1. Answer (1)O
CH3Carbanion becomes aromatic.
2. Answer (1)O
H C2 H
O
H C2 HCarbanion stabilize by resonance.
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3. Answer (1)
H C C OH3
H
NaOI
COONa
+ CHI3
C21. 1. Answer (3)NH2
HNO
H SO3
2 4
NH2
NO2
+
NH2
NO2M
(47%)P
(51%)Metal is formed due to formation of anilinium ion as
intermediate because of basic nature of aniline.
2. Answer (2)The carbonyl group in electron withdrawing making
the amide less basic than the amine.
3. Answer (2)The substitution is para, so the amide must be
ortho-para directing. The best explanation for the lack ofortho
substitution is steric hinderance.
C22. 1. Answer (3)
I
Ag O2
N(CH )3 3
OH
+ NMe + H O3 2NMe3
2. Answer (2)
CH NH2 23CH I3
CH N2
Me
Me
Me
I
Ag O2
+ Me N 3 + H O2
3. Answer (4)
N Me
nBu Et
OH CH = CH22
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Organic Chemistry Success Magnet (Solutions)
C23. 1. Answer (1)Due to ortho effect.
2. Answer (2)Conjugate base stabilise more than other site.
3. Answer (2)Conjugate pair stabilise by resonance.
C24. 1. Answer (3)
CH NH2 2A B
O
due to ring expansion.
2. Answer (2)Phenyl group migrated over to N atom.
3. Answer (2)R N = C = O is common in Schmidt and Hoffmann
reaction.
C25. 1. Answer (3)The lone pair of electrons on the nitrogen of
the amine attacks the electrophillic carbon of acid.
2. Answer (1)Hbonding in Nylon.
3. Answer (4)It is prepared by condensation of adipic acid and
hexamethylene diamine.
C26. 1. Answer (1)
C N
O H H bonding
2. Answer (2)Teflon is polymer of CF2 = CF2.
3. Answer (3)Polyacrylonitrile
H HC = CH CN
H H( C C ) H CN
n
C27. 1. Answer (2)
62
8.703.30pH =+=
2. Answer (4)Amino acids show lowest solubility at isoelectric
point since there is highest concentration of the dipolar ion.
3. Answer (4)Since it is basic amino acid with 2NH2 group, So pH
must be greater than 7.
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C28. 1. Answer (2)Shortening of carbon chain of one carbon.
2. Answer (3)Name of reaction Ruff degradation.
3. Answer (4)(A) C = NOH, (B) Pentyacetate(C) Cyanide, (D)
Aldopentose
C29. Starch(C H O )6 10 5 n
200 250C
(C H O )6 10 6 n
H O2 C H O12 22 11H O2 C H O6 12 6
conc.H SO2 4
(A)Dextrin
(C) (D)(E)
Black
1. Answer (2)2. Answer (1)3. Answer (1)
C30. 1. Answer (3)Elimination dominates over substitution.
2. Answer (4)A C3H6B CH CH CH3 3
BrC CH CH CH3 3
OH3. Answer (3)
Substitution product
Cl OH OH
C31. 1. Answer (1)
R C R
OH
OH
(O)R C R
O
2. Answer (4)All of these contain Cr+6
PCCN
HCrO3Cl
3. Answer (3)nfactor = 6
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Organic Chemistry Success Magnet (Solutions)
C32. 1. Answer (3)
C H C C H + MgX2 5 2 5 C H2 5
OH O3
+
C H C C H2 5 2 5
OH
C H2 5
H O2
C H C = C CH2 5 3C H2 5
C H C = O + CH C H2 5 3
O
C H2 5
H
2. Answer (3)Less hindered more reactive
3. Answer (3)C = O C
sp2 sp3
C33. 1. Answer (3)CH3
H SO2 4
CH3
SO H3 PCl5
CH3
SO Cl2 NH3
CH3
SO NH2 2
COOH
SO NH2 2
CO NH
SO2
CH3
SO H ( P)3
H SO2 4
2. Answer (1)CH3 CH3 CH3 CH3
SO H3
PCl5
SO Cl2
NH3
SO NH2 2
(i) ClOH(ii) NaOCl
SO NNa2+
Cl
3. Answer (4)
CH3
SO H3
PCl5
CH3
SO Cl2
C H OH6 5
CH3
SO OC H2 6 5ester
H O218
CH3
SO OH218
+
OH
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C34. 1. Answer (2)Rearrangement of carbocation.
2. Answer (3)Intermediate 2 carbocation of AIntermediate 1
carbocation of B
3. Answer (3)
OHCH NH2 2
HNOHN
2
2
+
OH
CH2ring expansion
OH OH O
C35. 1. Answer (2)
CH C CH2 3
O
CH C H2&
O
2 carbanion 4 product2. Answer (4)
Carbanion stabilize by resonance.3. Answer (4)
(1) & (2) option no H atom(3) option give chloroform
reaction.
Section - D : Assertion - Reason Type1. Answer (1)
That acid is easily decarboxylate which produce most stable
carbanion.
2. Answer (1)Due to I effect producing group.
3. Answer (1)
=+
Major23
Minor23 CHOCHHCOCH
Stabilize by resonance, the second structure has octets on all
atoms and an additional bond.
4. Answer (3)
HONO + H SO2 2 4 H O NO + HSO2 4
Hbase acid
5. Answer (2)
HOX + H + H O X (acid) H
H O + X2electrophile
6. Answer (2)The bond energy of allylic carbon hydrogen bond in
propene is less than the bond energy of benzylic carbon-hydrogen
bond in toluene.
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Organic Chemistry Success Magnet (Solutions)
7. Answer (3)Hyperconjugation increases stability of carbocation
and free radical, not carbanion.
8. Answer (1)N does not have vacant d-orbital for expansion of
octet. Maximum co-valency of N is 4.
9. Answer (3)Tartaric acid isomers = 3.d, l and meso form.
10. Answer (2)
O
NH2
H O/H or OH2+
OH
NH2
(1-5 migration of H atom).
11. Answer (3)1 is R, II S.
12. Answer (1)
C H
ClH
O
Cl OH
H-bonding
Cl C
13. Answer (2)
No. of isomers 21n
n 22
.
14. Answer (3)Double bond generating geometrical isomers and
chiral centre give optical isomers.
15. Answer (2)Unimolecular elimination.
16. Answer (4)A CHCl3 is more acidic than CHF3 because Cl C3 is
less basic than F C3 because fluorine can dispersecharge only by an
inductive effect while Cl disperse charge by inductive effect as
well as p p bondingdelocalisation.
17. Answer (1)Due to steric crowding electrons pair does not
undergo resonance.
18. Answer (4)Triplet carbene is more stable than singlet
carbene.
19. Answer (1)
Angle strain of cyclopentane = 75.021085.109
=
(negligible).
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20. Answer (1)
CH C3CH3O CH3
CH C3
CH3
O CH3(resonance)
21. Answer (2)Vinyl chloride does not form vinyl carbocation
with anhy. AlCl3.
22. Answer (3)Pyridine is less reactive than benzene towards EAS
creating at the position due to - I effect of the N-atom while
inpyrrole the non-bonding pair on nitrogen is part of aromatic
rextet.
23. Answer (1)Number of hyperconjugative structure
stability.
24. Answer (3)Acidic hydrogen is present in 1-alkynes but not in
alkyne-2.
25. Answer (2)Due to (1 4) position of H which causes hinderance
in boat form.
26. Answer (1)O H
H
O H
HH
H
H-bonding
27. Answer (4)Ethylene is more reactive than acetylene towards
electrophilic addition reaction.
28. Answer (4)Rate of nitration of C6H6 = rate of nitration of
C6D6.
29. Answer (1)
C C
Br+
Intermediate ionbromonium ion
30. Answer (2)Correct R Mesotartaric acid has molecular
symmetry.
31. Answer (4)
CH C C H3 H C C ==== C H3 CH C == C H3Hg+2 Hg Hg+
OH H
Hg+2CH C == CH3 2
OHOH H (enol)
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32. Answer (2)(II) Does not react with NH3 since double bond is
not polarised by an electron withdrawing group.
(I) Form product (III) Via H N C C C == C3 +CH3 H
CH3 H
O
CH3
intermediate.
33. Answer (3)Alkyne gives carbonyl compounds
R C C H (i) H O(BH )22 3THF(ii) H O OH2 2/
R CH C H2O
34. Answer (4)It involves formations of vinyl carbocation, which
is unstable reaction does not takes place.
35. Answer (1)
OH
ringexpansion
H+ H+
36. Answer (3)Order of boiling pointStraight chain > branched
chain of isomeric hydrocarbon.
37. Answer (1)Removal of Cl is easy due to presence of NO2
group.
38. Answer (4)Anti-Markownikov product.
39. Answer (4)Chloroform is heavier than water.
40. Answer (1)Racemic mixture is obtained due to walden
inversion.
41. Answer (4)C2H5Br + AgCN C2H5CN (major product)
42. Answer (2)Aniline behaves as Lewis base for anhydrous
AlCl3.
43. Answer (3)Addition of HBr to 2-pentene give meso
product.
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44. Answer (3)2 carbocation converts to 3 carbocation.
45. Answer (1)HCN is weak acid where as HI is strong.
46. Answer (3)Presence of electron withdrawing group at ortho
position of Aryl halide increases nucleophilic substitution.
47. Answer (3)Correct reason :- In aryl halides electron density
at the ring decreases due to electron withdrawing effect ofhalogen
atom.
48. Answer (1)
R O R + H
+ R OH R
49. Answer (1)Phenoxide ion stabilized by resonance.
50. Answer (2)Correct reason :- Formation of stable intermediate
benzyl carbocation.
51. Answer (3)Correct reason :- Intramolecular H-bonding
52. Answer (1)
OH
H
Benzene(aromatic)
53. Answer (2)Correct reason :- OH group of salicyladehyde is
less reactive due to presence of intramolecular H-bonding.
54. Answer (3)Correct reason :- Rearrangement of carbocation
form stable saytzeff product.
55. Answer (2)Intramolecular aldol reaction.
56. Answer (1)+I effect of CH3 group decrease dipole moment in
ketone.
57. Answer (4)It forms 1 nitro 2 propanol.
58. Answer (1)Cl is good leaving group than NH2 group.
59. Answer (4)p-chloro benzoic is more acidic than p-fluoro
benzoic acid due to more effective p-p bonding.
60. Answer (4)Guanidine is more basic due to stability of its
conjugate pair.
61. Answer (2)In strongly acidic medium protonation of hydoxyl
amine takes place.
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62. Answer (4)Chloro ethanoic acid has lowest value of pKa and
stronger acid.
63. Answer (3)Cross Cannizzaro is redox reaction.
64. Answer (2)Cl is good leaving group.
65. Answer (4)+I effect of methyl group.
66. Answer (3)It has 3 H atom.
67. Answer (1)HCHO has 2H for hydride shift and benzaldehyde is
resonance stabilised.
68. Answer (4)NaBH4 reduces only carbonyl group, not
carbon-carbon multiple bond.
69. Answer (3)OH abstracts the hydrogen from CH3NO2.
70. Answer (4)2, 2 dimethyl propanoic acid has no hydrogen does
not gives HVZ reaction.
71. Answer (1)Zn-Hg/conc. HCl form cyclo alkene.
72. Answer (1)Umbrella effect.
73. Answer (3)Pyrrole is a weak base. It is aromatic because the
non-bonding electrons on nitrogen are located in a p-orbital,where
they contribute to aromatic sextet.
74. Answer (2)Due to hindrance.
75. Answer (1)+I effect of alkyl group increase electron density
of N-atom.
76. Answer (4)Gabriel phthalimide is used for preparation of 1
amine in pure state.
77. Answer (2)The intermediate is stabilized by delocalization
of negative charge on to the electronegative ion.
78. Answer (2)
NH2C
NH2
NH
H2NC
NH2
NH2H O2 (is stabilised by resonance)
79. Answer (4)In alkaline medium with ZnNaOH, nitrobenzene gives
hydroazobenzene.
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80. Answer (1)CCl bond in chlorobenzene is resonance stabilised,
it has partial double bond character.
81. Answer (2)Amides are less basic than primary amines because
lone pair of electrons on nitrogen atom in amides
isdelocalised.
R C NH2
O
R C = NH2
O +
82. Answer (4)Carbylamine reaction is applicable to primary
amines.
83. Answer (1)
N N+
similarly on the three rings.
84. Answer (1)In acidic medium, aniline is converted to
anilinium ion which does not couple.
85. Answer (1)Carboxypeptidase is an exopeptidase as it breaks
the peptide chain at terminal ends. Carboxy peptidase
cleavescarboxy - terminal amino acids having aromatic or branched
aliphatic side chains.
86. Answer (3)Nylon is a polyamide of hexamethylene diamine and
adipic acid.
87. Answer (1)Bakelite becomes hard on heating, hardening is due
to formation of extensive cross-links between differentpolymer
chains to give a three dimensional network solid.
88. Answer (1)
Carbocation formed from styrene )CHHCHC( 356 +
is more stable than that formed from propene.
89. Answer (4)Glycine does not contain a chiral centre.
90. Answer (3)Teflon is fully fluoroninated polymer.
91. Answer (4)Nylon is a monomer of hexamethylene diammine and
adipic acid.
92. Answer (3)Polybutadiene is chain growth polymer.
93. Answer (3)Both glucose and fructose reduces Tollens
reagent.
94. Answer (3)Cn(H2O)n General formula.
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Organic Chemistry Success Magnet (Solutions)
95. Answer (1)On hydrolysis sucrose give unequal amount of
glucose and fructose which changes sign of rotation.
96. Answer (2)Enzymes are generally made up of proteins.
97. Answer (2)Fructose is monosaccharide carbohydrate which do
not undergo hydrolysis.
98. Answer (2)Existence of both cationic (NH3+) species and
anionic species (COO).
99. Answer (2)Starch is polymer of glucose.
100. Answer (4)-D-glucose and -D-glucose are anomers.
Section - E : Matrix-Match Type
1. Answer - A(p, q, s), B(p, q, r, s), C(p, q, r, s), D(p, q, r,
s)
OH
can react with Na, NaOH and NaNH2 but not with NaHCO3. While
others react with Na, NaOH, NaNH2
and NaHCO3.
2. Answer - A(p, r, s), B(q, r), C(q, r), D(p, r, s)If two (+I)
or two (I) effect groups are present on doubly bonded carbon atom
then dipole moment (cis > trans),Melting point (trans > cis)
and boiling point (cis > trans). If one (+I) and on (I) effect
group is present on bothboubly bonded carbon atom then dipole
moment (trans > cis) and melting point (trans > cis) but
boiling point(cis > trans).
3. Answer - A(q, r), B(p, q), C(r, q), D(q, r, s)
(A)
Aromatic and resonance stabilised (4n + 2 = 6)
(B) CH C C = C3CH3
CH3
CH3
H
+1 groupresonance
CH C = C C3
CH3 CH3
CH3H
(C) Aromatic and resonance stabilised (4n + 2 = 6)
(D)N O
OCH2
Aromatic resonance and I effect of NO2 group.
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4. Answer - A(p, r), B(p, q), C(s), D(r)Friedel Crafts
reaction
CH Cl3anhyd. AlCl3 CH3
CH3
Reimer Tiemann reactionOH
CHCl , KOH3
OHCHO
salicylaldehydeIntermediate Carbene, an electrophile attack on
electron rich ringAldol condensation
CH C H3
OOH CH C H2
O
carbanion
CH C H3
OCH C H2
O
CH CH CH CHO3 2
OH
Acid catalysed hydration Carbocation is intermediate
5. Answer - A(r), B(p), C(q), D(q, s)Chlorination in presence of
h is a free radical reaction.
Bromination of alkene proceeds through cyclic transition state
and undergoes anti addition.
Hydration involves carbocationic mechanism, carbocation formed
in this case will not undergo rearrangement.
Elimination (E1) proceeds through carbocation, which has
tendency of rearrangement.6. Answer - A(q, r, s), B(q), C(p, s),
D(s)
(A)H
Very acidic hydrogen so reacts with base it is allylic hydrogen
which will be subslituted by
Cl2/h. Presence of double bond will make it react with Br2
water.
(B) Acidic hydrogen HOCHH 2
(C) Undergoes nitration with (HNO3 + H2SO4) as well as free
radical addition with Cl2/h.
(D) free radical substitution with Cl2/h.7. Answer - A(s), B(q,
r, s), C(p, s), D(q, r, s)
(A) Cis elimination of meso compound produces trans isomer
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(B) Elimination is possible from non-stereo centre. So, is a
stereospecific reaction and a stereoselectivereactions and
undergoes antielimination.
BrH
C
Br H
C
HH
H
HHH
(C) Carbocation would be formed, reactant is optically active
product will show geometrical isomer hence thereaction is
stereoselective.
8. Answer - A(p, q, r), B(p, s), C(p, q, r), D(p, q, r)(A)
Dehydration (elimination of H2O) proceeds through carbocationic
intermediate which will undergo
rearrangement.(B) Elimination of HF from fluoroalkanes yields
Hoffmann product(C) Same as (1)(D) Elimination of HCl proceeds
through carbocation intermediate.
9. Answer - A(p, q, s), B(p, r, s), C(p, q, s), D(p, r, s)Non
terminal alkyne can form trans-alkene with Na/Liq.NH3 and does not
react with ammonical AgNO3 whileterminal alkynes do not form
trans-alkene with Na/Liq.NH3 but reacts with ammonical AgNO3.
10. Answer - A(p, q, r, s), B(r), C(p, q, r), D(r)(A) CH4 + Cl2
h CH3Cl + CH2Cl2 + CHCl3 + CCl4.(B) CHCl3 form carbene (:CCl2) with
KOH.(C) CH3Cl, CH2Cl2 and CHCl3 have dipole moment but CCl4 has
zero dipole moment.
(D)O
reacts with Cl2 and NaOH to give haloform reaction and produce
chloroform.
11. Answer - A(p, r, s), B(p, s), C(q), D(p)(A) Dehydration in
(A) will follow E1 mechanism, would be brought by H+, reaction will
proceed through
carbocation, which can undergo rearrangement.(B) Same as (A) but
carbocation will not undergo rearrangement, since it is benzylic
carbon.(C) Follows E2 mechanism.(D) E1 mechanism follows
carbocationic rearrangement.
12. Answer - A(p, s), B(p, r), C(r), D(q)(A) CH3 C C H HBO
OH2 CH3 CH = CH OH ationtautomeris CH3 CH2CHO
(B) CH3 C CH OMDM CH C = CH3 2OH
ationtautomeris CH C CH3 3
O
(C) Oxymercuration demercuration involves addition of H2O giving
Markownikovs product.(D) Oxo process.
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13. Answer - A(s), B(p), C(r), D(q)
Cl % = %54.201002.29
610058.34316
16==
+++
Cl % = %12.172.29
1005=
Cl % = %0.261002.29
6.7=
Cl % = %27.101002.293
=
14. Answer - A(p, r), B(p, r), C(q, s), D(s)(A) Alc. KOH will
bring elimination, reaction but the carbocation formed is stable so
reaction will proceed
through E1.
(B) Same.(C) Sterically less hindered alkyl halide so will
undergo E2 mechanism.(D) E1CB.
15. Answer - A(p, r, s), B(p), C(q), D(r, s)Carbocation is
stabilised by electron releasing group whereas it is destabilised
by electron withdrawing group.
16. Answer - A(p, q, s), B(q), C(p, r, s), D(p, q, s)
(A) C CH3OH
HH+ CH CH3
CH
= CH2
Elimination by E1 mechanism.
(B) 2 alkyl halides are more prone to elimination in presence of
sterically hindered base.(C) Carbocation formed is aromatic so
reaction proceeds through SN1.(D) 3 alcohol will undergo E1
mechanism in presence of H+.
17. Answer - A(q, r), B(q, r), C(p), D(s)(A) Friedel Craft
alkylation and a electrophilic substitution.(B) Friedel Craft
acylation and a electrophilic substitution.(C) Only possibility for
chlorobenzene is SN with KOH, though that also is not feasible.(D)
Chlorination in presence of h is free radical reaction.
18. Answer - A(r), B(q), C(p), D(s)Tertiary alkyl halide gives
E1 reaction while secondary alkyl halide gives E2 reaction. If two
basic groups arepresent then elimination through E1CB.
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19. Answer - A(q), B(s), C(p), D(r)
(A)Cl
reacts with aqueous KOH gives SN2 reaction.
(B)Cl
reacts with alcoholic KOH gives E2 reaction and converted into
alkene.
(C)Cl
reacts with H2O gives SN1 reaction and form tertiary
alcohol.
(D)OH
reacts with H+ and on heating gives E1 reaction due to formation
of tertiary carbocation.
20. Answer - A(r), B(p, r, s),
C(q, r), D(p, r)(D) Reactant molecule undergoes anti-elimination
but as the product cant exhibit stereoisomerism thats why
reaction is not stereospecific.
21. Answer - A(p), B(s),
C(q), D(q, r)(A) Picric acid give CO2 with NaHCO3.(B) Tertiary
alcohol gives white turbidity within few seconds.(C) 2 alcohol
gives white turbidity after 8 10 minutes. Ethanol gives iodoform
test.(D) Ethanol give iodoform test and evolve H2 with sodium
metal.
22. Answer - A(p, q, r), B(p, r),
C(p, s), D(p, r)(A) Stable carbocation intermediate due to
rearrangement and stable Saytzeff product.(B) 3 stable
carbocation.(C) E2 elimination.(D) Unimolecular elimination.
23. Answer - A(q, s), B(p, r, s), C(q), D(p, r)
(A)OCH3
OCH3
OCH3
3HI OH
OH
OH
+ 3CH I3
So, product reacts with Na and CH3I is one of the product.
(B)OCH3
OCH3
OCH36HI
II
I+ 3CH I + 3H O3 2
(C) OPh
OPh
OPh
3HI
II
I3PhOH +
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(D)CH OH2CHOHCH OH2
5HICH3CH ICH3 2 Alkyl halide
24. Answer - A(q), B(r), C(p, r), D(p)(A) In aldol condensation,
carbanion is formed as an intermediate.(B) In the formation of
Grignard reagent, free redical is formed as an intermediate.(C) In
the Ist step free radical is formed as an intermediate while in
IInd step carbocation is formed as an
intermediate.
(D) In dehydration, carbocation is formed as an intermediate.25.
Answer - A(p, q), B(p, r), C(q, s), D(r, s)
(A)O
+ C = OH
HOH
O
OH
(B)O
+C H
O
OH
O
CH
(C) C = OH
H+ CH C H3
OHO
CH C CH C H3 2
H O
OH
(D) CO
H+ CH C H3
C = C C H
H
H OO
26. Answer - A(p, q, r, s), B(p, s), C(q, s), D(q, s)Phenol
gives alkoxy benzene with Friedel Craft reaction, chloroform gives
carbyl amine, phosgene gas andReimer Tiemann reaction.
27. Answer - A(r), B(s), C(p), D(q)Based on data and acidic
strength.
28. Answer - A(q, s, r), B(q, r, s), C(p, q, r, s), D(q, r,
s)(A) A gives ketone with H3O+ and HgSO4 and by HBO also undergo
ozonolysis to form bicarbonyl compounds(B) Alkene give ozonolysis
and HBO reaction
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(C)OH
PCC
O
OH Zn/H O2+ O3
OH
OO
OHHBO
OH
OH
(D) O3Zn/H O2OO
29. Answer - A(q, r), B(p, r), C(s), D(p, s)
CH3 C
O
, gives haloform test.CHO gives positive Tollens test.OH and
COOH, gives NaHCO3 test.
30. Answer - A(p, q, r, s), B(p, q, r, s), C(q, r), D(p, s)Acid
with CaO/, acyl chloride with (CH3)2Cd and alkyl cyanide with
Grignard salt give ketone.
31. Answer - A(r), B(p, q, r), C(r, s), D(r, s)Cannizzaro
reaction is responded by aldehydes containing no H.Aldol
condensation is responded by carbonyl compounds containing acidic
hydrogen. Refomatsky reaction isresponded by aldehydes only Tollens
reagent can oxidise aldehydes only.
32. Answer - A(p, q, r, s), B(p, r), C(p, q, r, s), D(r)-keto
acids can be decarboxylated by heating.
33. Answer - A(r), B(s), C(p), D(q)
(A)
COCl
H O3+
COOH
(B)COCl
NH3
C NH2
(Amide)
O
(C)H OH2
CH3COOH ester
(D)
COCl
RCOOH Anhydride
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34. Answer - A(r, s), B(p, r), C(p, q), D(r, s)(A) Hofmann
bromamide(B) Mannic condensation(C) Aldol condensation(D) Schemidt
reaction
35. Answer - A(p, q, r, s), B(r, s), C(s), D(p, q, r, s)RMgX
reacts with acidic hydrogen.
36. Answer - A(p, r), B(p, q), C(p, q, r), D(s)(A) 2 amine gives
insoluble material with Hinsberg and yellow oily layer of
p-nitrosoamine.(B) 1 amine gives Hinsberg test and alcohol give red
colour with Victor Meyer.(C) 1 amine Hinsberg
NO2 Red colour with Victor Meyer.
(D) Only aldehyde give silver mirror with Tollens reagent37.
Answer - A(p, q), B(q), C(s), D(p, r, s)
Amide can give amine with reduction as well as Br2/KOH
38. Answer - A(p, r), B(q, r), C(p, s), D(q, r)
Only 3 N is present in benzene ring with sp2 hybridisation
whereas 1 and 2 are non-benzenoid heteroaromatic and 4 N is
aliphatic.
39. Answer - A(q), B(p, q, s), C(p, r), D(p, s)
(A) NCl
OHNOH
H+N
O H O
Ag+
N
+
(B)
NH2
HNO2
N2
OH
2 alcohol gives blue colourwith Victor Meyer
HClring expansion
(C) No ring expansion since CH2
is stable than
CH2OH gives red colour with
Victor Meyer.
(D)O N2 NH2 O N2 OH
gives Victor Meyer test
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40. Answer - A(p, q, r, s), B(p, r, s), C(r), D(q, r, s)(A)
Protein Polymer of amino acid formed by condensation of -amino
acids and carboxylic acid having
amide linkage.
(B) Nylon66 Condensation polymer having 2 monomeric units
forming N C O
H
bond.
(C) Buna N It is a copolymer of butadiene and vinyl cyanide.(D)
PHBV Poly hydroxy butyrate CO hydroxy valerate is a copolymer of
3-hydroxy butanoic acid
and 3-hydroxy pentanoic acid in which the monomer are connected
by ester linkages.
CH CH CH COOH + CH CH CH CH COOH3 2 3 2 2
OH
OH
PHBV
41. Answer - A(p), B(p), C(q, s), D(q, r)(A) Buna-S is a
co-polymer of butadiene and styrene. It is an addition polymer.(B)
Polythene is an additional polymer and is obtained by polymerizing
ethylene.(C) Nylon 6, 6 is obtained by condensation polymerization
of hexamethylene diamine and adipic acid.(D) Terylene is also
called a polyester as it contain ester group. It is a condensation
polymer.
42. Answer - A(q, p), B(q), C(r, s), D(s)(A) Ethylene glycol is
a monomor used in formation of Terylene and Glyptal.(B)
Terephthalic acid is a monomer of Terylene.(C) Formaldehyde is a
monomer used in formation of Bakelite as well as malmac.(D) Phenol
is a monomer used in formation of Bakelite.
43. Answer - A(q), B(q), C(q, s), D(p, r, s)(A) Nylon 6 Formed
by condensation of monomer.(B) Glyptal It is a polyester formed by
ethylene glycol and phthalic acid.(C) Nylon 2, 6 It is a
biodegradable, condensation polymer.(D) Cellulose A polymer of
-glucose units joined together by a glycosidic linkage is natural
occur polymer
and is biodegradable.44. Answer - A(p, q, s), B(p, r, s), C(s),
D(s)
(A) Glucose is monosaccharides with 5 chiral carbon and D
glucose differ in C1 configuration.(B) Glucose and mannose has same
molecular formula C6H12O6 but differ in configuration at 1-carbon
only.(C) Glucose and fructose are monosaccharides.(D) Ribose and
glucose are monosaccharides.
45. Answer - A(p, q, r, s), B(p, q, s), C(p, s), D(s)(A) Glucose
and mannose are monosaccharide both reducing sugar and are
C2-epimers.(B) Mannose and galactose are monosaccharides, reducing
sugar.(C) Glucose and fructose are monosaccharides and reducing
sugar.(D) Lactose and maltose are reducing sugar.
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46. Answer - A(p), B(p, q), C(r, s), D(p, q)(A) Cellulose is
polymer composed of D-glucose units which are joined by -glycosidic
linkages.(B) Proteins are nitrogeneous polymeric substance.(C)
Lipids are esters of long chain fatty acids and alcohols and
richest source of energy stored in the living
bodies.(D) Nucleic acids are a group of high molecular mass
biomolecules which are present in all living cells in
format nucleoproteins. Nucleoproteins are made up of proteins
and natural polymers.
47. Answer - A(q, r, s), B(p, r, s), C(q, r, s), D(q)Maltose,
sucrose and lactose are disaccharides having glycosidic linkage and
sucrose invert its configurationafter hydrolysis. Fructose is
reducing monosachharide sugar.
48. Answer - A(q, r), B(p, q), C(q, r), D(q, s)Glycine and
alanine are amino acid form Zwitter ion. Protein has amide linkage
form hydrogen bonds.
DNA Nucleotide containing deoxy ribose sugar.
Section - F : Subjective Type
1. (i)
O
I Highly strainedring
O
The positive charge is resonance stabilised
O
O
II
O
C
Greater charge separation in I assigns it a higher dipole moment
than II.
(ii)NH
NH
In N H , no such resonance is possible. Hence, electron density
is more over nitrogen atom.
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(iii)I
is aromatic as it involves cyclic delocalization and follows
Huckels (4n + 2) electron rule.
II is not aromatic and hence is less stable.
2. (i)O
CH2H3C
OCH3H3C
OOH
Both OC and OH groups are in trans position and so no hydrogen
bonding exists and can easily
tautomerise and it is more polar
OH
OH C3 CH3 CH3H C3
O O
(I) (II)Here, intramolecular hydrogen bonding exists and is less
polar. Thus compound (I) is more acidicthan (II).
(ii)N O
is basic but NH
Ois not because in the first case, the lone pair on nitrogen do
not
participate in resonance with O since it will generate a double
bond on bridge head position while
in the second case it does participate in resonance, decreasing
its basicity.
NH
O
NH
O+
(iii) OOH
is more acidic than O OH
because its conjugate base is more stabilised.
OO
O
O
O
O
O
OO
O
3. (a) In compound (i) and (iii), there is plane of symmetry
passing through the compounds (molecular plane).Therefore, they are
optically inactive. Compound (ii) does not have plane of symmetry
as the two phenylrings are not in the same plane. One of the ring
rotates about CC bond axis because of bulkysubstituents at ortho o
positions of two adjacent phenyl rings and the two rings are
perpendicular toeach other. So (ii) is optically active.
(b) Compound (i) is optically active because there is no plane
of symmetry which can cut the molecule intotwo equal halves.
Compound (ii) also does not have plane of symmetry so, it is
optically active.Compound (iii) is optically inactive because of
the presence of centre of symmetry.
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4. (i) N OHN
O
(ii) Ph C C CH3
O
O
Ph C C === CH2
O O
H
(iii)O OH O
O
(iv) N == OHO N OHO
(v)OO
O
OHHO
OH
(vi)O OH
(vii) NH
N
(viii) CH NO3 2 CH N2 == OH
O
(ix) CO
CH3CH3
H3C CH3 C
OH
CH3CH3
H3C CH3
(x) H C3 CH3O
H C3 CH3OH
(xi) C = OH C3
H C3C OH
H C2
H C3
(xii)CH3CH3
H3CO
H3C has no -H atom; hence keto-enol tautomerism is not
possible.
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5. (i)
OHBr Br
Br
(ii)OH
(iii)Cl CH3
N
O
H
(iv)
OEtOEt
NO2NO2
Cl
6.